Question

line integral

Answer 1

F=<X,Y> and C is a triangle with vertices (0,-4), (0,4), (3,0) oriented counter clockwise

eval \[\int\limits_{C}^{}F.dr\]

Answer 2

Do you know Green's theorem by any chance?

Answer 3

not much of it

Answer 4

It's probably not needed here, just wondering. Do you know about conservative vector fields?

Answer 5

yes

Answer 6

Great. If the path is closed and the partial derivative of \(F\) are equal, what can you say about the line integral?

Answer 7

its conservative

Answer 8

Right, the vector field is conservative, but what does that mean for the integral itself?

Answer 9

would it equal something like F(b)-F(a) or something?

Answer 10

Exactly, and if the path is closed, then \(a=b\).

Answer 11

but im confused sicne its multivariable

Answer 12

Actually, slight correction. Not \(F(b)-F(a)\), but\(f(b)-f(a)\) where \(f(x,y)\) is the potential function. The potential function is such that \(\nabla f(x,y)=F\) and \(F\) is the vector field.

Answer 13

so what, i integrate <x,y> to get <1/2x^2,1/2y^2>?

Answer 14

That's the basic idea, but there are some details to keep track of. You have \(F=(x,y)\) a conservative vector field, so you can find its potential function \(f\).
\[\frac{\partial f}{\partial x}=x~~\implies~~f=\int x\,\mathrm{d}x=\frac{1}{2}x^2+g(y)\]where \(g(y)\) is a function that depends only on \(y\).

Differentiating with respect to \(y\), you have
\[\frac{\partial f}{\partial y}=y=g'(y)~~\implies~~g(y)=\frac{1}{2}y^2+C\]So, the potential function is \(f(x,y)=\dfrac{1}{2}x^2+\dfrac{1}{2}y^2+C\).

Now,
\[\int_C \vec{F}\bullet\,\mathrm{d}\vec{r}=\int_C \nabla f(x,y)\bullet\,\mathrm{d}\vec{r}=f(\vec{b})-f(\vec{a})\]where \(\vec{a}=(x_0,y_0)\) is the starting point on the path and \(\vec{b}=(x_1,y_1)\) is the end of the path.

If the path is closed, then \(\vec{a}=\vec{b}\).

Answer 15

in that case its 0?

Answer 16

ya it is, wow

Answer 17

Yup